Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
// テ・ツ淞コテヲツ慊ャテ」ツδ?」ツδウテ」ツδ療」ツδャテ」ツδシテ」ツδ?

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define int long long int

template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}

typedef pair<int, int> pii;
typedef long long ll;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
constexpr ll INF = 1001001001001001LL;
constexpr ll MOD = 1000000007LL;

int N, M, s, t;
vector<int> G[100010];
int ca[100010], cb[100010];

vector<int> dijkstra(int s) {
    vector<int> ret(N, INF);
    ret[s] = 0;

    // dist, point
    priority_queue< pii, vector<pii>, greater<pii> > q;
    q.push(pii(0, s));

    while(!q.empty()) {
        pii t = q.top(); q.pop();
        int cost = t.first, cur = t.second;
        for(int to : G[cur]) {
            if(ret[to] > ret[cur] + 1) {
                ret[to] = ret[cur] + 1;
                q.push(pii(ret[to], to));
            }
        }
    }

    return ret;
}

signed main() {
    cin >> N >> M >> s >> t; s--; t--;
    rep(i,0,M) {
        int x, y; cin >> x >> y; x--; y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }

    vector<int> da = dijkstra(s), db = dijkstra(t);
    rep(i,0,N) if(da[i] != INF) ca[ da[i] ]++;
    rep(i,0,N) if(db[i] != INF) cb[ db[i] ]++;

    
    int dist = da[t], ans = 0;
    repq(i,0,dist-2) {
        int j = dist - 2 - i;
        ans += ca[i] * cb[j];
    }
    cout << ans << endl;
    return 0;
}